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The existence and multiplicity of positive solutions of nonlinear sixth-order boundary value problem with three variable coefficients

Wanjun Li

Author affiliations

Department of Mathematics, Longdong University, Qingyang 745000, Gansu, P. R. China

Citation and License

Boundary Value Problems 2012, 2012:22  doi:10.1186/1687-2770-2012-22

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/22


Received:22 November 2011
Accepted:22 February 2012
Published:22 February 2012

© 2012 Li; licensee Springer.

This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this article, we discuss the existence and multiplicity of positive solutions for the sixth-order boundary value problem with three variable parameters as follows:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M1">View MathML</a>

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous. The proof of our main result is based upon spectral theory of operators and fixed point theorem in cone.

Keywords:
sixth-order differential equation; positive solution; fixed point theorem; spectral theory of operators

1 Introduction

In this article, we study the existence and multiplicity of positive solution for the following nonlinear sixth-order boundary value problem (BVP for short) with three variable parameters

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M2">View MathML</a>

(1.1)

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous.

In recent years, BVPs for sixth-order ordinary differential equations have been studied extensively, see [1-7] and the references therein. For example, Tersian and Chaparova [1] have studied the existence of positive solutions for the following systems (1.2):

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M3">View MathML</a>

(1.2)

where A, B, and C are some given real constants and f(x, u) is a continuous function on R2, is motivated by the study for stationary solutions of the sixth-order parabolic differential equations

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M4">View MathML</a>

This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and solid state. When f(x, u) = u - u3, it was studied by Gardner and Jones [2] as well as by Caginalp and Fife [3]. In [1], existence of nontrivial solutions for (1.2) is proved using a minimization theorem and a multiplicity result using Clarks theorem when C = 1 and f(x, u) = u3. The authors have studied also the homoclinic solutions for (1.2) when C = -1 and f(x, u) = -a(x)u|u|σ, where a(x) is a positive periodic function and σ is a positive constant by the mountain-pass theorem of Brezis-Nirenberg and concentration-compactness arguments. In [4], by variational tools, including two Brezis-Nirenbergs linking theorems, Gyulov et al. have studied the existence and multiplicity of nontrivial solutions of BVP (1.2).

Recently, in [5], the existence and multiplicity of positive solutions of sixth-order BVP with three parameters

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M5">View MathML</a>

(1.3)

has been studied under the hypothesis of

(A1) f : [0,1] × [0, ∞) → [0. ∞) is continuous.

(A2) α, β, γ R and under the condition of satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M6">View MathML</a>

the existence and multiplicity for positive solution of BVP (1.3) are established by using fixed point index theory. In this article, we consider more general BVP (1.1), based upon spectral theory of operators and fixed point theorem in cone, we will establish the existence and multiplicity positive solution of BVP (1.1) and extend the result of [5] under appropriate conditions. Our ideas mainly come from [5,8-10].

We list the following conditions for convenience:

(H1) f : [0,1] × [0, +∞) → [0. +∞) is continuous.

(H2) A(t), B(t), C(t) ∈ C[0,1], α = min0≤t≤1 A(t), β = min0≤t≤1 B(t), γ = min0≤t≤1 C(t), and satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M7">View MathML</a>

Let Y = C[0,1], Y+ = {u Y : u(t) ≥ 0, t ∈ [0,1]}. It is well known that Y is a Banach space equipped with the norm ||u||0 = sup0≤t≤1 |u(t)|, u Y. Set X = { u C4[0,1] : u(0) = u(1) = u''(0) = u''(1) = 0}, then X also is a Banach space equipped with the norm ||u||X = max {||u(t)||0, ||u"(t)||0, ||u(4)(t)||0}. If u C4[0,1] ∩ C6(0,1) fulfils BVP (1.1), then we call u is a solution of BVP (1.1). If u is a solution of BVP (1.1), and u(t) > 0, t ∈ (0, 1), then we say u is a positive solution of BVP (1.1).

2 Preliminaries

In this section, we will make some preliminaries which are needed to show our main results.

Lemma 2.1. Let u X, then ||u||0 ≤ ||u"||0 ≤ ||u(4)||0 ≤ ||u||X.

Proof. The proof is similar to the Lemma 1 in [8], so we omit it. □

Lemma 2.2. [5] Let λ1, λ2, and λ3 be the roots of the polynomial P (λ) = λ3 + γλ2 - βλ + α. Suppose that condition (H2) holds, then λ1, λ2, and λ3 are real and greater than -π2.

Note : Based on Lemma 2.2, it is easy to learn that when the three parameters satisfy the condition of (H2), they satisfy the condition of non-resonance.

Let Gi(t, s)(i = 1, 2, 3) be the Green's function of the linear BVP

-u"(t) + λiu(t) = 0, u(0) = u(1) = 0,

Lemma 2.3. [10]Gi(t, s)(i = 1, 2, 3) has the following properties

(c1) Gi(t, s) > 0, ∀t, s ∈ (0, 1).

(c2) Gi(t, s) <CiGi(s, s), ∀t, s ∈ [0,1], in which Ci > 0 is constant.

(c3) Gi(t, s) ≥ δiGi(t, t)Gi(s, s), ∀t, s ∈ [0,1], in which δi > 0 is constant.

We set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M8">View MathML</a>

(2.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M9">View MathML</a>

(2.2)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M10">View MathML</a>

(2.3)

then starting from Lemma 2.3 we know Mi, mi, Cij > 0.

For any h Y, take into consideration of linear BVP:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M11">View MathML</a>

(2.4)

where α, β, γ satisfy assumption (H2). Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M12">View MathML</a>

(2.5)

then for any h Y, the LBVP(2.4) has a unique solution u, which we denoted by Ah = u. The operator A can be expressed by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M13">View MathML</a>

(2.6)

Lemma 2.4. The linear operator A : Y X is completely continuous and ||A|| ≤ ϖ, where ϖ = |λ23|(C1C2C3M1M2M3|λ3|+C1C2M1M2)+| λ2λ3|(C1C2C3M1M2M3+C1M1).

Proof. It is easy to show that the operator A : Y X is linear operator. ∀h Y, u = Ah X, u(0) = u(1) = u"(0) = u"(1) = u(4)(0) = u(4)(1) = 0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M14">View MathML</a> that is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M15">View MathML</a>

(2.7)

by (2.5) and (2.7), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M16">View MathML</a>

and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M17">View MathML</a> so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M18">View MathML</a>

(2.8)

By (2.6), for any t ∈ [0,1], we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M19">View MathML</a>

(2.9)

Again, let ω = -u" + λ3u, then ω(0) = ω(1) = ω"(0) = ω"(1) = 0, by (2,5), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M20">View MathML</a>

(2.10)

Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M21">View MathML</a> that is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M22">View MathML</a>

(2.11)

So

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M23">View MathML</a>

(2.12)

Based on (2.8), (2.9), and (2.12), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M24">View MathML</a>

(2.13)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M25">View MathML</a>

(2.14)

So, ||u(4)(t)|| ≤ ϖ||h||0, by Lemma2.1, ||u||X ϖ||h||0, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M26">View MathML</a>

(2.15)

so A is continuous, and ||A|| ≤ ϖ.

Next, we will show that A is compact with respect to the norm ||·||X on X.

Suppose {hn}(n = 1, 2, . . .) an arbitrary bounded sequence in Y, then there exists K0 >0 such that ||hn||0 K0, n = 1, 2, . . . . Let un = Ahn, 1, 2, ...By (2.8), ∀t1, t2 ∈ [0, 1], t1 < t2, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M27">View MathML</a>

Because Gi(t, s)(i = 1, 2, 3) is uniform continuity on [0,1] × [0,1], based on the above demonstration, it is easy to proof that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M28">View MathML</a> is equicontinuous on [0,1]. From (2.15), we know ||u||0, ||u"||0, ||u(4)||0 ≤ ||u||X ϖ||hn||0 ϖK0, so <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M29">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M30">View MathML</a> are relatively compact in R. Based on Lemma 1.2.7 in [11], we know <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M31">View MathML</a> is the relatively compact in X, so A is compact operator. □

The main tools of this article are the following well-known fixed point index theorems.

Let E be a Banach Space and K E be a closed convex cone in E. Assume that Ω is a bounded open subset of E with boundary ∂Ω, and K ∩ Ω ≠ ∅. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M32">View MathML</a> be a completely continuous mapping. If Au u for every u K ∩ ∂Ω, then the fixed point index i(A, K ∩ Ω, K) is well defined. We have that if i(A, K ∩ Ω, K) ≠0, then A has a fixed point in K ∩ Ω.

Let Kr = {u K |||u|| <r} and ∂Kr = {u K |||u|| <r} for every r >0.

Lemma 2.5. [12] Let A : K K be a completely continuous mapping. If μAu u for every u ∈ ∂Kr and 0 < μ ≤ 1, then i(A, Kr, K) = 1.

Lemma 2.6. [12] Let A : K K be a completely continuous mapping. Suppose that the following two conditions are satisfied:

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M33">View MathML</a>

(ii) μAu u for every u ∈ ∂Kr and μ ≥ 1,

then i(A, Kr, K) = 0.

Lemma 2.7. [12] Let X be a Banach space, and let K X be a cone in X. For p >0, define <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M34">View MathML</a>. Assume that A : Kp K is a completely continuous mapping such that Au u for every u ∈ ∂Kp = {u K|||u|| = p}.

(i) If ||u|| ≤ ||Au||, for every u ∈ ∂Kp, then i(A, Kp, K) = 0.

(ii) If ||u|| ||Au||, for every u ∈ ∂Kp, then i(A, Kp, K) = 1.

3 Main results

We bring in following notations in this section:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M35">View MathML</a>

Suppose that:

(H3) L = ϖK <1, where ϖ is defined as in (2.14).

Theorem 3.1. Assume that (H1)-(H3) hold, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M36">View MathML</a> then in each of the following cases:

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M37">View MathML</a> (ii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M38">View MathML</a>the BVP (1.1) has at least one positive solution.

Proof. ∀h Y, consider the LBVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M39">View MathML</a>

(3.1)

It is easy to prove (3.1) is equivalent to the following BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M40">View MathML</a>

(3.2)

where Gv := (C(t) - γ)v(4) - (B(t)- β)v" + (A(t) - α)v, ∀v X. Obviously, the operator G : X Y is linear, and ∀v X, t ∈ [0,1], we have |Gv(t)| ≤ K ||v||X. Hence ||Gv||0 K ||v||X, and so ||G|| ≤ K. On the other hand, u C4[0,1]⋂C6(0,1), t ∈ [0,1] is a solution of (3.2) iff u X satisfies u = A(Gu + h), i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M41">View MathML</a>

(3.3)

Owing to G : X Y and A : Y X, the operator I - AG maps X into Y. From A ϖ (by Lemma 2.4) together with ||G|| ≤ K and condition (H3), applying operator spectral theorem, we have that the operator (I - AG)-1 exists and is bounded. Let H = (I - AG)-1A, then (3.3) is equivalent to u = Hh. By the Neumann expansion formula, H can be expressed by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M42">View MathML</a>

(3.4)

The complete continuity of A with the continuity of (I - AG)-1 yields that the operator H : Y X is completely continuous. If we restrict H : Y+ Y, ∀h Y+ and mark u = Ah, then u X Y+. Based on equation (2.8), (2.11) and Lemma 2.4, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M43">View MathML</a>

by b(t) ≥ (λ2 + λ3)c(t) and<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M44">View MathML</a>, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M45">View MathML</a>

Hence

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M46">View MathML</a>

(3.5)

and so (AG)(Ah)(t) = A(GAh)(t) ≥ 0, ∀t ∈ [0,1]. Suppose that ∀h Y+, (AG)k (Ah)(t) ≥ 0, ∀t ∈ [0,1]. For any h Y+, let h1 = GAh, by (3.5) we have h1 Y+, and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M47">View MathML</a>

Thus by induction it follows that ∀n ≥ 1, ∀h Y+, (AG)n (Ah)(t) ≥ 0, ∀t ∈ [0,1]. By (3.4), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M48">View MathML</a>

(3.6)

So H : Y+ Y+ X.

On the other hand, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M49">View MathML</a>

(3.7)

So the following inequalities hold

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M50">View MathML</a>

(3.8)

For any u Y+, define Fu = f(t, u). Based on condition (H1), it is easy to show F : Y+ Y+ is continuous. By (3.1)-(3.3), It is easy to see that u C4[0,1] ∩ C6(0, 1) is a positive solution of BVP (1.1) iff u Y+ is a nonzero solution of an operator equation as follows

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M51">View MathML</a>

(3.9)

Let Q = HF. Obviously, Q : Y+ Y+ is completely continuous. We next show that the operator Q has at least one nonzero fixed point in Y+.

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M52">View MathML</a>

In which

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M53">View MathML</a>

(3.10)

Here M1 and M2 can be defined as that in (2.1), C12 and C23 can be defined as that in (2.2), Ci,δi(i = 1, 2, 3) can be defined as that in Lemma 2.3. It is easy to prove that P is a cone in Y. We will prove QP P next.

For any u P, let h = Fu, then h Y+. By (3.6) and Lemma 2.3, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M54">View MathML</a>

By Lemma 2.3, for all u P, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M55">View MathML</a>

And accordingly we have<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M56">View MathML</a>, that is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M57">View MathML</a>

(3.11)

By using (c3) in Lemma 2.3, (3.8) and (3.11), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M58">View MathML</a>

So<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M59">View MathML</a>. Thus QP P.

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M60">View MathML</a>

(3.12)

in which m1 can be defined as that in (2.1). It's easy to prove

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M61">View MathML</a>

(3.13)

Case (i), since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M62">View MathML</a>, there exist ε >0 and r0 >0 such that f(t, x) ≥ (Γ + ε)x, 0 ≤ t ≤ 1, 0 < × r0. Let r ∈ (0, r0) and Ωr = {u P | ||u||0 r}, then for every u ∈ ∂Ωr, we have ||u||0 = r, 0 < u(t) ≤ r, t ∈ (0, 1), and so f(t, u(t)) ≥ (Γ + ε)u(t), t ∈ (0,1). By (3.13), it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M63">View MathML</a>

(3.14)

From (3.6) and (3.14), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M64">View MathML</a>

Therefore, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M65">View MathML</a>. Now we shall prove ∀u ∈ ∂Ωr, μ ≥ 1, μQu u. In fact, suppose the contrary, then there exist <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M66">View MathML</a>, and μ0 ≥ 1 such that μ0Qu0 = u0. By (3.6), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M67">View MathML</a>. Let ω0 = AFu0, then u0 ω0 and ω0(t) satisfies BVP (2.4) with h = Fu0. Hence

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M68">View MathML</a>

(3.15)

After multiplying the two sides of the first equation in (3.15) by sin &#960t and integrating on [0,1], we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M69">View MathML</a>

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M70">View MathML</a>

(3.16)

Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M71">View MathML</a>, so <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M72">View MathML</a> and we see that Γ + ε < Γ, which is a contradiction. Then based on Lemma 2.6, we come to

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M73">View MathML</a>

(3.17)

On the other hand, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M74">View MathML</a>, there exist ε ∈ (0, (1 - L)Γ) and R0 >0 such that f(t, x) ≤ [(1-L)Γ - ε] x, 0 ≤ t ≤ 1, x >R0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M75">View MathML</a>. Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M76">View MathML</a>

We choose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M77">View MathML</a> and let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M78">View MathML</a> Next we prove ∀u ∈ ∂ΩR, μ ≥ 1, μu Qu. Assume on the contrary that ∃μ0 ≥ 1, u0 ∈ ∂ΩR, such that μ0u0 = Qu0. Let ω1 = AFu0, by (3.6), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M79">View MathML</a> and ω1(t) satisfies BVP (2.4) with h = Fu0. Similarly to (3.16), we can prove

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M80">View MathML</a>

(3.18)

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M81">View MathML</a>

(3.19)

Thus, by (3.19), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M82">View MathML</a> which is contradictory with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M83">View MathML</a>

Then by Lemma 2.5 we know

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M84">View MathML</a>

(3.20)

Now, by the additivity of fixed point index, combine (3.17) and (3.20) to conclude that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M85">View MathML</a>

Therefore Q has a fixed point in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M86">View MathML</a> which is the positive solution of BVP (1.1).

Case (ii), since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M87">View MathML</a> based on the definition of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M88">View MathML</a> we may choose ε >0 and ω >0, so that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M89">View MathML</a>

(3.21)

Let r ∈ (0, ω), we now prove that μQu u for every u ∈ ∂Ωr, and 0 < μ ≤ 1. In fact, suppose the contrary, then there exist u0 ∈ ∂Ωr, and 0 < μ0 ≤ 1 such that μ0Qu0 = u0. Let ω2 = AFu0, by (3.6), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M90">View MathML</a> and ω2(t) satisfies BVP (2.4) with h = Fu0. Similarly to (3.18), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M91">View MathML</a>

(3.22)

Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M92">View MathML</a> We see that (1 - L)Γ ≤ (1 - L)Γ - ε, which is a contradiction. By Lemma 2.5, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M93">View MathML</a>

(3.23)

On the other hand, because <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M94">View MathML</a> there exist ε ∈ (0, Γ) and H >0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M95">View MathML</a>

(3.24)

Let C = max0≤t≤1,0≤xH|f(t,x) - (Γ + ε)x| + 1, then it is clear that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M96">View MathML</a>

(3.25)

Choose R > R0 = max {H/ρ, ω}, ∀u ∈ ∂ΩR. By (3.13) and (3.25), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M97">View MathML</a>

And so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M98">View MathML</a>

(3.26)

From (3.6) and (3.26), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M99">View MathML</a>

from which we see that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M100">View MathML</a> namely the hypotheses (i) of Lemma 2.6 holds. Next, we show that if R is large enough, then μQu u for any u ∈ ∂ΩR and μ ≥ 1. In fact, suppose the contrary, then there exist u0 ∈ ∂ΩR and μ0 ≥ 1 such that μ0Qu0 = u0, then by (3.6), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M101">View MathML</a>Let ω0 = AFu0, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M102">View MathML</a> and ω0 satisfies BVP (2.4), in which h = Fu0, consequently,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M103">View MathML</a>

(3.27)

After multiplying the two sides of the first equation in (3.27) by sin πt and integrating on [0,1], we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M104">View MathML</a>

Consequently, we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M105">View MathML</a>

(3.28)

It's easy to prove that ω0(t), the solution of LBVF (3.27) satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M106">View MathML</a>

and accordingly,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M107">View MathML</a>

(3.29)

by (3.28), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M108">View MathML</a>

(3.30)

Consequently, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M109">View MathML</a>

We choose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M110">View MathML</a>then to any u ΩR, μ ≥ 1, there is always μQu u. Hence, hypothesis (ii) of Lemma 2.6 also holds. By Lemma 2.6, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M111">View MathML</a>

(3.31)

Now, by the additivity of fixed point index, combine (3.23) and (3.31) to conclude that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M112">View MathML</a>

Therefore, Q has a fixed poind in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M113">View MathML</a> which is the positive solution of BVP (1.1). The proof is completed. □

From Theorem 3.1, we immediately obtain the following.

Corollary 3.1. Assume (H1)-(H3) hold, then in each of the following cases:

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M114">View MathML</a> (ii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M115">View MathML</a>

the BVP (1.1) has at least one positive solution.

4 Multiple solutions

Next, we study the multiplicity of positive solutions of BVP (1.1) and assume in this section that

(H4) there is a p >0 such that 0 ≤ u p and 0 ≤ t ≤ 1 imply f(t, u) < ηp, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M116">View MathML</a>

(H5) there is a p >0 such that σp u p and 0 ≤ t ≤ 1 imply f (t, u) ≥ λp, where<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M117">View MathML</a>. Here, σ can be defined as (3.10).

Theorem 4.1. Assume (H1)-(H4) hold. If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M118">View MathML</a> and<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M119">View MathML</a>, then BVP (1.1) has at least two positive solution u1 and u2 such that 0 ≤ ||u1||0 p ≤ ||u2||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < r0 < p < R1 <+∞, such that 0 < r < r0 implies i(Q, Ωr, P) = 0 and R R1 implies i(Q, ΩR, P) = 0.

Next we prove i(Q, Ωp, P) = 1 if (H4) is satisfied. In fact, for every u ∈ ∂Ωp, based on the preceding definition of Q we come to

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M120">View MathML</a>

Consequently,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M121">View MathML</a>

Therefore, by (ii) of Lemma 2.7 we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M122">View MathML</a>

(4.1)

Combined with (3.17), (3.31), and (4.1), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M123">View MathML</a>

Therefore, Q has fixed points u1 and u2 in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M124">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M125">View MathML</a> respectively, which

means that u1(t) and u2(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p ≤ ||u2||0. The proof is completed. □

Theorem 4.2. Assume (H1)-(H3) and (H5) can be established, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M126">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M127">View MathML</a>, then BVP (1.1) has at least two positive solution u1 and u2 such that 0 ≤ ||u1||0 p ≤ ||u2||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < ω < p < R2 <+ ∞, such that 0 < r < ω implies i(Q, Ωr, P) = 1 and R R2 implies i(Q, ΩR, P) = 1.

We now prove that i(Q, Ωp, P) = 0 if (H5) is satisfied. In fact, for every u Ωp, by (3.13) we come to ρp ρ||u||0 u(t) ≤ ||u||0 = p, t ∈ [1/4, 3/4], accordingly, by (H5), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M128">View MathML</a>

from the proof of (ii) of Theorem 3.1, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M129">View MathML</a>

Therefore, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M130">View MathML</a>according to (i) of Lemma 2.7, we come to

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M131">View MathML</a>

(4.2)

Combined with (3.20), (3.23), and (4.2), there exist

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M132">View MathML</a>

Therefore, Q has fixed points u1 and u2 in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M133">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M134">View MathML</a> respectively, which means that u1(t) and u2(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p ≤ ||u2||0. The proof is completed. □

Theorem 4.3. Assume that (H1)-(H3) hold. If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M135">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M136">View MathML</a>, and there exists p2 > p1 >0 that satisfies

(i) f(t, u) < ηp1 if 0 ≤ t ≤ 1 and 0 ≤ u p1,

(ii) f(t, u) ≥ λp2 if 0 ≤ t ≤ 1 and σp2 u p2,

where η, σ, λ are just as the above, then BVP (1.1) has at least three positive solutions u1, u2, and u3 such that 0 ≤ ||u1||0 p1 ≤ ||u2||0 p2 ≤ ||u3||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < r0 < p1 < p2 < R3 <+∞, such that 0 < r < r0 implies i(Q, Ωr, P) = 0 and R R3 implies i(Q, ΩR, P) = 1.

From the proof of Theorems 4.1 and 4.2, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M137">View MathML</a>,<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M138">View MathML</a>. Combining the four afore-mentioned equations, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M139">View MathML</a>

Therefore, Q has fixed points u1, u2 and u3 in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M140">View MathML</a>and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/22/mathml/M141">View MathML</a>, respectively, which means that u1(t), u2(t) and u3(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p1 ≤ ||u2||0 p2 ≤ ||u3||0. The proof is completed. □

Competing interests

The author declares that they have no competing interests.

Authors' contributions

WL conceived of the study, and participated in its design and coordination. The author read and approved the final manuscript.

Acknowledgements

The author is very grateful to the anonymous referees for their valuable suggestions, and to be sponsored by the Tutorial Scientific Research Program Foundation of Education Department of Gansu Province P.R.China(1110-05).

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